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Math Help - Graphing

  1. #1
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    Graphing

    I need to graph the function s(x)=200x+1500/0.02x^2+5

    I cant figure out how to graph this function...I'm used to functions without (/) in it!
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  2. #2
    MHF Contributor Amer's Avatar
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    find critical point f'(x)=0 so you can find the increasing and decreasing intervals that may help and find the concavity of the curve you should first find f''(x)=0 after finding the points (critical points,and the points where f''(x)=0 )but them in the coordinates and find the point where the curve intersect x-axes (y=0) and y-axis (x=0) .....
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  3. #3
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    Quote Originally Posted by lisa1984wilson View Post
    I need to graph the function s(x)=200x+1500/0.02x^2+5

    I cant figure out how to graph this function...I'm used to functions without (/) in it!
    The / is an operator used for division.
    Akin to + for addition and - for subtraction
    and * or x for multiplication.

    Is you equation this?

     y = 200x + \frac {1500}{0.02 x^2} + 5

    If yes then a few of the x,y values

    y.value x.value
    75205 1
    19155 2
    8398.3 3
    2755 10
    4192.5 20

    Generate the x,y values from you function and then plot the data.
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  4. #4
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    The equation is set up like a big fraction not division
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  5. #5
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    Quote Originally Posted by lisa1984wilson View Post
    I need to graph the function s(x)=200x+1500/0.02x^2+5

    I cant figure out how to graph this function...I'm used to functions without (/) in it!
    next time, use the proper grouping symbols to make your expression clear ...

    s(x) = (200x+1500)/(0.02x^2+5)

    note that the numerator has degree less than the denominator ...

    s(x) is asymptotic to y = 0, the x-axis.

    s(x) has a single root at x = -7.5 and is negative-valued for x < -7.5 ; positive valued for x > -7.5

    y-intercept is (0,300)

    since the denominator is always positive, there are no vertical asymptotes or removable discontinuities.

    s'(x) = \frac{-10000(x-10)(x+25)}{(x^2+250)^2}

    at x = 10, s(x) has a relative maximum because s'(x) changes sign from (+) to (-)

    at x = -25, s(x) has a relative minimum because s'(x) changes sign from (-) to (+)

    s''(x) will tell you about inflection points and concavity.
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